$f$ is bijective, but is not the identity function $f(x) = x$.

Question

For the problem:

Let $A = \{1, 2, 3, 4, 5, 6\}$. Give an example of a function $f : A → A$ such that $f$ is bijective, but is not the identity function $f(x) = x$.

Is this valid example:

$f=\{(1,2), (2,1), (3,3), (4,4), (5,5), (6,6)\}$.

?

If yes, why is this not an identity function? My textbook doesn't say a lot about identity functions.

It's in your first box: the identity map $f$ satisfies $f(x)=x$ for all $x$. — Randall, Oct 22 '19 at 13:09
The identity function on the set is $I = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}$. Your function is not equal to $I$, since it contains $(1, 2)$, but $I$ does not. — Theo Bendit, Oct 22 '19 at 13:13

score 2 · Answer 1 · answered Oct 22 '19 at 13:19

Let $X$ be a finite set. A permutation of $X$ is a bijection $\sigma : X \to X$. Now let $X = I_{m} := \{1,...,m\}$ with $m \ge 2$ an integer. A permutation $\tau: I_{m} \to I_{m}$ is called a transposition if there exists positive integers $a \neq b$ in $I_{m}$ such that $\tau(a) = b$, $\tau(b) = a$ and $\tau(k) = k$ for $k \not\in \{a,b\}$. Thus, in your case you have $A = I_{6}$ and all possible transpositions (and permutations) that are not the identity map satisfy the requirements. For example, take $\tau :I_{6} \to I_{6}$ to be the map $\tau(1) = 2$, $\tau(2) =1 $ and $\tau(k) = k$ if $k \not\in \{1,2\}$

$f$ is bijective, but is not the identity function $f(x) = x$.

1 Answers1